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Liono4ka [1.6K]
4 years ago
10

( n + -2 ) x ( n - -4 ) = 27 What are the two n's?

Mathematics
2 answers:
Flauer [41]4 years ago
5 0

Simplify n + -2 to n - 2

(n - 2)(n + 4) = 27

Expand

n^2 + 4n - 2n - 8 = 27

Simplify n^2 + 4n - 2n - 8 to n^2 + 2n - 8

n^2 + 2n - 8 = 27

Move all terms to one side

n^2 + 2n - 8 - 27 = 0

Simplify n^2 + 2n - 8 - 27 to n^2 + 2n - 35

n^2 + 2n - 35 = 0

Factor n^2 + 2n - 35

(n - 5)(n + 7) = 0

Solve for n

<u>n = 5, -7</u>

storchak [24]4 years ago
5 0

Answer:

9 and 3

Step-by-step explanation:

27/3=9

There are 2 different ways to solve this equation.

Side note: subtracting a negative (4- -3) turns the equation into addition (4+3) and adding a negative (4+ -3) turns the equation into subtraction (4-3)

So the original equation: (n + -2)x(n - -4)=27

Turns into this: (n-2)x(n+4)=27

Way 1: (11-2)x(-1+4)=27

9x3=27

Way 2: (5-2)x(5+4)=27

3x9=27

Hope this helped! :)

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Reika [66]

Answer:

m=0

Step-by-step explanation:

if that's not how you write it then try y=0x

6 0
3 years ago
Pls help!!!!Shawn used 15 centimeters of tape to wrap 5 presents . How much tape will Shawn need I all if he has to wrap 13 pres
Ilia_Sergeevich [38]

Answer:39

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15÷5=3

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3 0
4 years ago
I need help in this section.
Phoenix [80]

Answer:

\frac{1}{45}

Step-by-step explanation:

There are

2 yellow

3 magenta

5 blue

marbles in the bag.

So total of 2 + 3 + 5 = 10 marbles

We have to multiply the probability of yellow on first pick by the probability of yellow on second pick (WITHOUT REPLACEMENT).

We denote P(Y1) as probability of yellow on first draw &

P(Y2) as probability of yellow on second draw

Thus,

P(Y1) = 2/10 = 1/5  [since 2 yellow and 10 total marbles]

Now,

P(Y2) = 1/9  [since now 1 yellow is taken out and NOT replaced, so we have 1 yellow remaining and total 9 marbles]

Now, we multiply:

P(Y1) * P(Y2) = 1/5 * 1/9 = 1/45

The probability is  \frac{1}{45}

6 0
3 years ago
Suppose the value R(d) of d dollars in euros is given by R(d)-d The cost P(n) in dollars to purchase and ship n purses is given
S_A_V [24]

<u>Corrected Question</u>

Suppose the value R(d) of d dollars in euros is given by R(d)=\dfrac67 d. The cost in dollars to purchase and ship n purses is given by P(n)=66n+23. Write a formula for the cost, Q(n) in euros to purchase and ship n purses.

Answer:

Q(n)=\dfrac67(66n+23)

Step-by-step explanation:

The value R(d) of d dollars in euros is given by R(d)=\dfrac67 d

Therefore:

R(1)=\dfrac67

T$hat is, 1  dollars =\dfrac67$ euros

The cost P(n) in dollars to purchase and ship n purses is given by:

P(n) = 66n+23.

Therefore, the cost, Q(n) in euros to purchase and ship n purses

=\dfrac67 \cdot P(n)\\Q(n)=\dfrac67(66n+23)

8 0
3 years ago
Let f(x)=7x+4 find each value <br> A. f(x) when x = -8 <br> B. f(3) <br> C. x when f(x) = 67
maks197457 [2]

Answer:

f(x)=7x+4


When x=-8

f(-8)=7·(-8)+4=-56+4=-52


When x=3

f(3)=7·3+4=21+4=25


When f(x)=67, then :

7x+4=67

7x=63

x=9

8 0
2 years ago
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